1. Find oxidation state and d-electron count of each metal ion

• (i) $[FeF_6]^{3-}$

  Let oxidation state of Fe be $x$: $x + 6(-1) = -3 \Rightarrow x=+3$ So, $Fe^{3+}$ has configuration: $Fe: [Ar]3d^64s^2 \Rightarrow Fe^{3+}: 3d^5$

• (ii) $[Co(NH_3)_6]^{3+}$

  $NH_3$ is neutral, so: $x=+3$ Thus $Co^{3+}$: $Co: [Ar]3d^74s^2 \Rightarrow Co^{3+}: 3d^6$

• (iii) $[NiCl_4]^{2-}$

  Let oxidation state of Ni be $x$: $x+4(-1)=-2 \Rightarrow x=+2$ So $Ni^{2+}$: $Ni: [Ar]3d^84s^2 \Rightarrow Ni^{2+}: 3d^8$

• (iv) $[Cu(NH_3)_4]^{2+}$

  $NH_3$ is neutral, so Cu is $+2$: $Cu^{2+}: 3d^9$

2. Determine spin state / number of unpaired electrons

Magnetic moment (spin only) depends on number of unpaired electrons: $\mu = \sqrt{n(n+2)}\;BM$ where $n$ = number of unpaired electrons.

So we only need to compare $n$.

(i) $[FeF_6]^{3-}$

• Octahedral complex
• $F^-$ is a weak field ligand
• Hence high-spin $d^5$

For high-spin $d^5$: $n=5$

(ii) $[Co(NH_3)_6]^{3+}$

• Octahedral complex
• $Co^{3+}$ is $d^6$
• With $NH_3$ and high oxidation state $+3$, this complex is low spin

Low-spin $d^6$ octahedral: $t_{2g}^6 e_g^0$ So, $n=0$

(iii) $[NiCl_4]^{2-}$

• Tetrahedral complex
• $Cl^-$ is weak field
• Tetrahedral complexes are generally high spin
• For tetrahedral $d^8$, configuration gives: $n=2$

(iv) $[Cu(NH_3)_4]^{2+}$

• $Cu^{2+}$ is $d^9$
• A $d^9$ system always has: $n=1$

3. Compare magnetic moments

Since $\mu$ increases with number of unpaired electrons, decreasing magnetic moment means decreasing $n$:

• (i): $n=5$
• (iii): $n=2$
• (iv): $n=1$
• (ii): $n=0$

Therefore, correct decreasing order is: $(i) > (iii) > (iv) > (ii)$

4. Match with options

This corresponds to: $\boxed{\text{C}}$

5. Comparison with stored correct answer

Stored correct answer = C

Our derived answer = C

So they agree.